java 中的欧拉计划 #8
Project Euler #8 in java
这里列出了挑战:
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
找出 1000 位数字中相邻的 13 个数字的乘积最大。这个产品的价值是多少?
我编写的代码适用于给定的 4 位数示例,但不适用于 13 位数。我怀疑存在某种类型的数据溢出,但我不确定。我超级低效的代码如下。
public class Euler8 {
public static void main(String[]args){
String num = "/*number listed above*/";
int n1, n2, n3, n4, n5, n6, n7, n8, n9, n10, n11, n12, n13;
long sum=0, newSum;
for(int n=0; n<=987; n++){
n1=Character.getNumericValue(num.charAt(n));
n2=Character.getNumericValue(num.charAt(n+1));
n3=Character.getNumericValue(num.charAt(n+2));
n4=Character.getNumericValue(num.charAt(n+3));
n5=Character.getNumericValue(num.charAt(n+4));
n6=Character.getNumericValue(num.charAt(n+5));
n7=Character.getNumericValue(num.charAt(n+6));
n8=Character.getNumericValue(num.charAt(n+7));
n9=Character.getNumericValue(num.charAt(n+8));
n10=Character.getNumericValue(num.charAt(n+9));
n11=Character.getNumericValue(num.charAt(n+10));
n12=Character.getNumericValue(num.charAt(n+11));
n13=Character.getNumericValue(num.charAt(n+12));
newSum= (long)(n1*n2*n3*n4*n5*n6*n7*n8*n9*n10*n11*n12*n13);
if(newSum>=sum)
sum=newSum;
}
System.out.println(sum);
}
}
我的代码输出这个数字:
2091059712
您的代码转换为 long
为时已晚:在执行转换时,乘法已在 32 位整数中完成,可以预见会导致溢出。
修改如下代码修复问题:
// newSum should be called newProd, because you use multiplication, not addition
newSum= ((long)n1)*n2*n3*n4*n5*n6*n7*n8*n9*n10*n11*n12*n13;
请注意,您的算法不是最有效的:如果您观察到位置 i+1
..i+13
的乘积可以从位置的乘积计算出来,那么您的算法可以快 13 倍i
..i+12
除以位置 i
的值并乘以位置 i+13
.
的值
当然你必须小心不要被零除。您可以通过观察任何时候遇到零来解决此问题,接下来的 13 个产品都将是零,因此您可以简单地跳过它们,然后继续下一个 "train" 个非零。
问题是 n1*n2*n3*n4*n5*n6*n7*n8*n9*n10*n11*n12*n13
溢出,因为:
- 都是
int
变量,
int
乘以 int
得到 int
.
long
的 typecase 应用于整个产品,(因此)发生得太晚,无法使用 long
算法完成计算并避免溢出问题。
该特定问题的简单解决方案是将 n
变量声明为 long
。 @dasblinkelights 的代码(转换 n1
)可能更快……但您需要对其进行基准测试才能确定。还有比这更重要的优化。
我得到的答案是 23514624000。这实际上是正确的。
public class LargestProduct {
public static void main(String[]args) {
String s="7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
long k=0,l=13,ans=1,n1,n2,n3,n4,n5,n6,n7,n8,n9,n10,n11,n12,n13;
long Max_Num=0;
String[] result=new String[1000];
result=s.split("");
for(int i=0;i<=987;i++) {
n1=Integer.parseInt(result[i]);
n2=Integer.parseInt(result[i+1]);
n3=Integer.parseInt(result[i+2]);
n4=Integer.parseInt(result[i+3]);
n5=Integer.parseInt(result[i+4]);
n6=Integer.parseInt(result[i+5]);
n7=Integer.parseInt(result[i+6]);
n8=Integer.parseInt(result[i+7]);
n9=Integer.parseInt(result[i+8]);
n10=Integer.parseInt(result[i+9]);
n11=Integer.parseInt(result[i+10]);
n12=Integer.parseInt(result[i+11]);
n13=Integer.parseInt(result[i+12]);
ans=n1*n2*n3*n4*n5*n6*n7*n8*n9*n10*n11*n12*n13;
Max_Num=Math.max(Max_Num, ans);
}
System.out.println(Max_Num);
}
}
这里列出了挑战:
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450
找出 1000 位数字中相邻的 13 个数字的乘积最大。这个产品的价值是多少?
我编写的代码适用于给定的 4 位数示例,但不适用于 13 位数。我怀疑存在某种类型的数据溢出,但我不确定。我超级低效的代码如下。
public class Euler8 {
public static void main(String[]args){
String num = "/*number listed above*/";
int n1, n2, n3, n4, n5, n6, n7, n8, n9, n10, n11, n12, n13;
long sum=0, newSum;
for(int n=0; n<=987; n++){
n1=Character.getNumericValue(num.charAt(n));
n2=Character.getNumericValue(num.charAt(n+1));
n3=Character.getNumericValue(num.charAt(n+2));
n4=Character.getNumericValue(num.charAt(n+3));
n5=Character.getNumericValue(num.charAt(n+4));
n6=Character.getNumericValue(num.charAt(n+5));
n7=Character.getNumericValue(num.charAt(n+6));
n8=Character.getNumericValue(num.charAt(n+7));
n9=Character.getNumericValue(num.charAt(n+8));
n10=Character.getNumericValue(num.charAt(n+9));
n11=Character.getNumericValue(num.charAt(n+10));
n12=Character.getNumericValue(num.charAt(n+11));
n13=Character.getNumericValue(num.charAt(n+12));
newSum= (long)(n1*n2*n3*n4*n5*n6*n7*n8*n9*n10*n11*n12*n13);
if(newSum>=sum)
sum=newSum;
}
System.out.println(sum);
}
}
我的代码输出这个数字:
2091059712
您的代码转换为 long
为时已晚:在执行转换时,乘法已在 32 位整数中完成,可以预见会导致溢出。
修改如下代码修复问题:
// newSum should be called newProd, because you use multiplication, not addition
newSum= ((long)n1)*n2*n3*n4*n5*n6*n7*n8*n9*n10*n11*n12*n13;
请注意,您的算法不是最有效的:如果您观察到位置 i+1
..i+13
的乘积可以从位置的乘积计算出来,那么您的算法可以快 13 倍i
..i+12
除以位置 i
的值并乘以位置 i+13
.
当然你必须小心不要被零除。您可以通过观察任何时候遇到零来解决此问题,接下来的 13 个产品都将是零,因此您可以简单地跳过它们,然后继续下一个 "train" 个非零。
问题是 n1*n2*n3*n4*n5*n6*n7*n8*n9*n10*n11*n12*n13
溢出,因为:
- 都是
int
变量, int
乘以int
得到int
.
long
的 typecase 应用于整个产品,(因此)发生得太晚,无法使用 long
算法完成计算并避免溢出问题。
该特定问题的简单解决方案是将 n
变量声明为 long
。 @dasblinkelights 的代码(转换 n1
)可能更快……但您需要对其进行基准测试才能确定。还有比这更重要的优化。
我得到的答案是 23514624000。这实际上是正确的。
public class LargestProduct {
public static void main(String[]args) {
String s="7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
long k=0,l=13,ans=1,n1,n2,n3,n4,n5,n6,n7,n8,n9,n10,n11,n12,n13;
long Max_Num=0;
String[] result=new String[1000];
result=s.split("");
for(int i=0;i<=987;i++) {
n1=Integer.parseInt(result[i]);
n2=Integer.parseInt(result[i+1]);
n3=Integer.parseInt(result[i+2]);
n4=Integer.parseInt(result[i+3]);
n5=Integer.parseInt(result[i+4]);
n6=Integer.parseInt(result[i+5]);
n7=Integer.parseInt(result[i+6]);
n8=Integer.parseInt(result[i+7]);
n9=Integer.parseInt(result[i+8]);
n10=Integer.parseInt(result[i+9]);
n11=Integer.parseInt(result[i+10]);
n12=Integer.parseInt(result[i+11]);
n13=Integer.parseInt(result[i+12]);
ans=n1*n2*n3*n4*n5*n6*n7*n8*n9*n10*n11*n12*n13;
Max_Num=Math.max(Max_Num, ans);
}
System.out.println(Max_Num);
}
}